The cable of a 3750 lb elevator in the figure below snaps when the elevator is at rest at the first floor so that the bottom is a distance d = 12.0 ft above a cushioning spring whose force constant is k = 10,000 lb/ft. A safety device clamps the guide rails, removing 1000 ft-lb of mechanical energy for each 1.00 ft that the elevator moves.(adsbygoogle = window.adsbygoogle || []).push({});

(a) Find the speed of the elevator just before it hits the spring. mgh-1000h=mv^2/2 - 3750(12)-1000(12) = (3750/32)v^/2

v= 23.7318

(b) Find the distance that the spring is compressed. I got 2.8587 ft for this, but apparently it's wrong. Here's what I did:

mv^2/2 + mgx - 1000x = 1/2kx^2

(3750/32)(23.7318^2)/2 + 3750x - 1000x = 10000x^2/2

33000 + 2750x = 5000x^2

x= 2.8587

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# Conservation of energy - the elevator question

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